Why Are There Two Possible Values of x in Similar Shapes Geometry Problems?

[Steve4Physics]

Other value of x = 14.5)

Well done! I hope you drew diagrams:

I found this one very tough, but then it was Higher tier (GCSE, UK), so it might be aimed at students who are hoping for an 8 or a 9. I put a lot of work in to get a 7 last year.

For those who don’t know, the UK GCSE exam' determines the ability/attainment-level of students, most commonly at around age 16. Grades 1 to 9 are awarded with 9 being the highest.

In math’s only around 5% of students achieve grade 9 (though, sadly, this figure varies regionally across the UK).

I'd guess that this question was written with the intention to differentiate (in the non-calculus sense!) the grade 9s. So the question is written with the expectation that (some) grade 9 candidates should be able to answer in full - but the large majority of candidates should not.

 

[hutchphd]

:furiously Googles "ansatz":

it has so much more pizzazz than "guess", don't you think....?

 

[Beyond3D]

...

 

[Beyond3D]

Homework Statement: See screenshot in main body
Relevant Equations: I don't know if there is a generic equation
Maybe something like AB(SF)=AC ?

View attachment 363169
Why are there two possible values of x?

Here's what I've done so far.

AE(SF)=AD
12(SF)=15
SF=15/12
=5/4

AB(SF)=AC
8(5/4)=AC
AC=10

AC-AB=x
x=10-8
x=2

A hint please, as to my next step

Let me help, it is a pretty basic geometry problem if you know the rules behind it
Assume that the triangles are congruent (scalable to eachother) and that the ratios between the sides remain constant
$$\ AB:AE=AC:AD$$
$$\frac{8cm}{12cm}=\frac{x+8cm}{15cm}$$
$$15\cdot\frac{8cm}{12cm}=15\cdot\frac{x+8cm}{15cm}$$
$$\frac{120}{12}=x+8$$
$$\ 10 = x+8$$
$$\ x=2$$

 

[DaveC426913]

Let me help, it is a pretty basic geometry problem if you know the rules behind it
Assume that the triangles are congruent (scalable to eachother) and that the ratios between the sides remain constant
$$\ AB:AE=AC:AD$$
$$\frac{8cm}{12cm}=\frac{x+8cm}{15cm}$$
$$15\cdot\frac{8cm}{12cm}=15\cdot\frac{x+8cm}{15cm}$$
$$\frac{120}{12}=x+8$$
$$\ 10 = x+8$$
$$\ x=2$$

OK, but that's only one of the two indicated solutions.

And you've based it on this assumption:
$$\ AB:AE=AC:AD$$

 

[Beyond3D]

OK, but that's only one of the two indicated solutions.

And you've based it on this assumption:
$$\ AB:AE=AC:AD$$

Yes, and that assumption is *heavily* implied graphically.

 

[DaveC426913]

Yes, and that assumption is *heavily* implied graphically.

1. In geometry, there are only three ways of determining that two lines are parallel: definition, postulate, and proof. Diagrams are not to be measured or eyeballed or heavily implied.
2. Never mind "implication" or "assumption"; it is explicitly stated in the problem that there are two solutions. (perhaps you have not fully read the assignment?)

 

[TensorCalculus]

Yes, and that assumption is *heavily* implied graphically.

Let's not start this argument all over again

Yes, maybe the diagram is misleading in order to push students to think harder (or, to make the question writer look smarter. We don't know.). But the problem says 2 solutions, and to even further assist the student in realising what the other answer is, they have asked you to state any assumptions that you have made in your workings.
The answer that the OP was struggling with (and has now found) was via not making that assumption.

 

[Beyond3D]

1. In geometry, there are only three ways of determining that two lines are parallel: definition, postulate, and proof. Diagrams are not to be measured or eyeballed or heavily implied.

2. Never mind "implication" or "assumption"; it is explicitly stated in the problem that there are two solutions. (perhaps you have not fully read the assignment?)

I know there are two solutions, but how would you get the second?

 

[TensorCalculus]

I know there are two solutions, but how would you get the second?

Do you want me to tell you or would you rather work it out yourself?
You've assumed that $AB:AE=AC:AD$ - in the other case that isn't true

 

[Beyond3D]

Do you want me to tell you or would you rather work it out yourself?
You've assumed that $AB:AE=AC:AD$ - in the other case that isn't true

Can you give me a little nudge, just a slight hint?

 

[TensorCalculus]

Similar triangles means that they have all the same angles. What's the other option for the triangles to be similar in this case, if that assumption isn't true?

Edit: if you want a little more help, what if $AE:AC = AB:AD$ ?

 

[Steve4Physics]

Can you give me a little nudge, just a slight hint?

It's already posted! See diagram in Post #91.

 

[TensorCalculus]

It's already posted! See diagram in Post #91.

Oh wow, I just realised, we let this thread drag on 104 messages... feels like it was only 10 messages long like 2 seconds ago .

 

[hutchphd]

Let me add to the clutter to re-emphasize:
If you make an assumption and it leads to a valid solution, that does not invalidate the solution!! It does not, however, validate the assumption.
I am beginning to love this question. Often I get to a solution by making a guess. One cannot be blinded into assuming this guess becomes truth. Therein lies the path to ruin. In this question, the presence of an "official" drawing is problematic. Perhaps the question should first ask for two drawings ? I don't really see a cleaner way to present this in a text.

 

[TensorCalculus]

What I really want to do now is find out who the writer of this exam question was and email them a link to this thread: just for fun!! Wonder what the reaction would be.

 

[Steve4Physics]

What I really want to do now is find out who the writer of this exam question was and email them a link to this thread: just for fun!! Wonder what the reaction would be.

Well done - you extended the thread to 4 'pages'! So I won't feel guilty about adding this...

$$\ AB:AE=AC:AD$$
$$\frac{8cm}{12cm}=\frac{x+8cm}{15cm}$$
$$15\cdot\frac{8cm}{12cm}=15\cdot\frac{x+8cm}{15cm}$$
$$\frac{120}{12}=x+8$$
$$\ 10 = x+8$$
$$\ x=2$$

FWIW, there's quicker method. One of the solutions requires BE to be parallel to CD (which is evident when you do a drawing). So you can use the intercept theorem and immediately write (with all values in cm):
$\frac x8 = \frac 3{12}$ so $x = 8 \times \frac 3{12} = 2$.
(But this is not applicable to the other solution.)

 

[fresh_42]

Oh wow, I just realised, we let this thread drag on 104 messages... feels like it was only 10 messages long like 2 seconds ago .

There is nothing that extends a thread more than two different opinions! I remember a seemingly endless thread about the existential question whether $0$ is real. Or start a thread by asking whether $\dfrac{1}{2}=\dfrac{2}{4}$ represents an equivalence relation or actual equality. It will be hilarious. And bloody.

 

[TensorCalculus]

There is nothing that extends a thread more than two different opinions! I remember a seemingly endless thread about the existential question whether $0$ is real. Or start a thread by asking whether $\dfrac{1}{2}=\dfrac{2}{4}$ represents an equivalence relation or actual equality. It will be hilarious. And bloody.

okay time to start a thread about 1/2 = 2/4

 

[paulb203]

What I really want to do now is find out who the writer of this exam question was and email them a link to this thread: just for fun!! Wonder what the reaction would be.

Maths Genie
https://www.mathsgenie.co.uk/resources/1hnov2017B.pdf
Q.22

 

[TensorCalculus]

Maths Genie
https://www.mathsgenie.co.uk/resources/1hnov2017B.pdf
Q.22

I'll have to hunt down the people who wrote the Edexcel gsce maths papers no?

 

[paulb203]

I'll have to hunt down the people who wrote the Edexcel gsce maths papers no?

Ah, I see what you mean. Yeah.

 

[Patxitxi]

Homework Statement: See screenshot in main body
Relevant Equations: I don't know if there is a generic equation
Maybe something like AB(SF)=AC ?

View attachment 363169
Why are there two possible values of x?

Here's what I've done so far.

AE(SF)=AD
12(SF)=15
SF=15/12
=5/4

AB(SF)=AC
8(5/4)=AC
AC=10

AC-AB=x
x=10-8
x=2

A hint please, as to my next step

A jocular note:
I didn't know that "similar" was a mathematical term.
Proportional would have appropriate.
Unless the intended effect was to turn Thales on his grave with this much ado about nothing.

 

[TensorCalculus]

Thanks to @pbuk I knew the origin of the question (edexcel GCSE maths papers) but sadly they don't reveal the names/contacts of the individual question writers... what does everyone think about emailing this address? General Edexcel contact:
examsofficers@pearson.com
I am very curious as to how they will react to our very... thorough analysis of their question :D

 

[fresh_42]

A jocular note:
I didn't know that "similar" was a mathematical term.
Proportional would have appropriate.
Unless the intended effect was to turn Thales on his grave with this much ado about nothing.

Proportional only refers to two quantities; similarity refers to all quantities, here three side lengths or three angles. It is a stronger condition that proportional would be.

 

[Muu9]

but sadly they don't reveal the names/contacts of the individual question writers

Probably for the best. I can only imagine them being signed up for tons of spam emails by chuffed students if their emails were public.

 

[TensorCalculus]

Probably for the best. I can only imagine them being signed up for tons of spam emails by chuffed students if their emails were public.

oh that makes sense.
I'm going to mail their main contact email anyway - just for fun! Maybe they will find it interesting to see such an opinionated converstaion about their question. They probably never expected people would think so hard about it haha.

Or maybe they will do what most do and ignore me. In that scenario, at least it was worth a try